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3 years ago

2NBr3 + 3NaOH – N2 + 3NaBr + 3HBrO

Chemistry

1 answer:

Damm [24]3 years ago

3 0

Answer:

50 formula units of NBr3 will yield 75 formula units of NaBr

57 formula units of NaOH will yield 57 formula units of NaBr

Explanation:

The equation of the reaction is ;

2NBr3 + 3NaOH – N2 + 3NaBr + 3HBrO

Number of formula units of NaBr formed when 50NBr3 reacts was obtained from;

2 formula units of NBr3 yields 3 formula units of NaBr

50 formula units of NBr3 will yield 50 ×3/2 = 75 formula units of NaBr

Hence 50 formula units of NBr3 will yield 75 formula units of NaBr

Then;

Since 3 formula units of NaOH yields 3 formula units of NaBr

57 formula units of NaOH will yied 57×3/3 = 57 formula units of NaBr

Hence 57 formula units of NaOH will yield 57 formula units of NaBr

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What is the total number of electrons in all s orbitals of a neutral atom of phosphorus?

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3 years ago

4. Al2O3 (s) + 6HCl (aq) → 2AlCl3 (aq) + 3H20(1) Find the mass of AlCl3 that is produced when 10.0 grams of Al2O3 react with 10.

slava [35]

Answer:

Are produced 12,1 g of AlCl₃ and 5,33 g of Al₂O₃ are left over

Explanation:

For the reaction:

Al₂O₃ (s) + 6 HCl (aq) → 2AlCl₃ (aq) + 3H₂0(l)

10,0g of Al₂O₃ are:

10,0g ₓ $\frac{1mol}{102g}$ = 0,0980 moles

And 10,0g of HCl are:

10,0 gₓ $\frac{1mol}{36,5g}$ = 0,274 moles

For a total reaction of 0,274 moles of HCl you need:

0,274× $\frac{1molesAl_{2}O_3}{6 mole HCl}$ = 0,0457 moles of Al₂O₃

Thus, limiting reactant is HCl

The grams produced of AlCl₃ are:

0,274 moles HCl × $\frac{2 moles AlCl_{3}}{6 moles HCl}$ × 133 $\frac{g}{mol}$ = 12,1 g of AlCl₃

The moles of Al₂O₃ that don't react are:

0,0980 moles - 0,0457 moles = 0,0523 moles

And its mass is:

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I hope it helps!

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4 years ago

Consider these three titrations: (i) the titration of 25.0 mL of a 0.100 M monoprotic weak acid with 0.100 M NaOH (ii) the titra

Setler79 [48]

Complete Question:

Consider these three titrations: (i) the titration of 25.0 mL of a 0.100 M monoprotic weak acid with 0.100 M NaOH (ii) the titration of 25.0 mL of a 0.100 M diprotic weak acid with 0.100 M NaOH (iii) the titration of 25.0 mL of a 0.100 M strong acid with 0.100 M NaOH. Which statement is most likely to be true?

(a) All three titrations require the same volume of NaOH to reach their first equivalence point.

(b) All three titrations have the same initial pH.

Answer:

(a) All three titrations require the same volume of NaOH to reach their first equivalence point.

Explanation:

(i) the titration of 25.0 mL of a 0.100 M monoprotic weak acid with 0.100 M NaOH

number of moles of acid = $\frac{25}{1000} dm^{3} * 0.1 M$ = 0.0025 moles

ii) the titration of 25.0 mL of a 0.100 M diprotic weak acid with 0.100 M NaOH

number of moles of acid = $\frac{25}{1000} dm^{3} * 0.1 M$ = 0.0025 moles

(iii) the titration of 25.0 mL of a 0.100 M strong acid with 0.100 M NaOH

number of moles of acid = $\frac{25}{1000} dm^{3} * 0.1 M$ = 0.0025 moles

Therefore, all the acids require the same number of moles of NaOH to reach their first equivalence points

Note that the concentration of the base NaOH are also the same, therefore the volume of NaOH required to reach equivalence point would also be the same for all the three titrations.

All three titrations don't have the same initial and equivalence point pH because they all have different acidic properties.

6 0

4 years ago

Read 2 more answers